Asked by Mary
A uranium nucleus (238U) moves along the positive ๐ฅ-axis with speed ๐ฃ = 5.0 ร 105 m/s when it decays
into an ๐ผ-particle (4He, helium nucleus) and a thorium nucleus (234Th). If 4He moves off at an angle of 20.2ยฐ above the ๐ฅ-axis with a speed ๐ฃ๐ป๐ = 1.4 ร 10^7 m/s, you will determine what is the recoil velocity, ๐ฃ , of 234Th and the angle below the ๐ฅ-axis. Assume the uranium-thorium-alpha system is isolated with not external forces acting on it and the masses of are A ร 1.673 ร 10^โ27 kg. This is a two-dimensional problem.
into an ๐ผ-particle (4He, helium nucleus) and a thorium nucleus (234Th). If 4He moves off at an angle of 20.2ยฐ above the ๐ฅ-axis with a speed ๐ฃ๐ป๐ = 1.4 ร 10^7 m/s, you will determine what is the recoil velocity, ๐ฃ , of 234Th and the angle below the ๐ฅ-axis. Assume the uranium-thorium-alpha system is isolated with not external forces acting on it and the masses of are A ร 1.673 ร 10^โ27 kg. This is a two-dimensional problem.
Answers
Answered by
Damon
You did not type clearly what the masses are
call mass U238 = mu
call mass Th = mt
call mass alpha = ma
speed of Th = st
speed of U = su
speed of alpha = 1.4*10^7
hopefully the Helium mass + the Thorium mass is the Uranium mass
then momentum is conserved
A is angle below x axis of Th
x direction
mu * 5*10^5 = ma * 1.4*10^7 * cos 20.2 + mt * st * cos A
y direction
0 = ma * 1.4*10^7 * sin 20.2 + mt* st * sin A
solve for the Th speed st and Angle A below x axis
call mass U238 = mu
call mass Th = mt
call mass alpha = ma
speed of Th = st
speed of U = su
speed of alpha = 1.4*10^7
hopefully the Helium mass + the Thorium mass is the Uranium mass
then momentum is conserved
A is angle below x axis of Th
x direction
mu * 5*10^5 = ma * 1.4*10^7 * cos 20.2 + mt * st * cos A
y direction
0 = ma * 1.4*10^7 * sin 20.2 + mt* st * sin A
solve for the Th speed st and Angle A below x axis